Question
part b: what is the margin of error for this interval?

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = $12.16 Population standard deviation = $\sigma$ =$1.87

Sample size = n =44

At 99% confidence level the z is ,

$\alpha$  = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z$\alpha$/2 = Z0.005 = 2.576

Margin of error = E = Z$\alpha$/2* ($\sigma$ /$\sqrt$n)

= 2.576 * ( 1.87/ $\sqrt$ 44)

= 0.7

At 99% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

12.16 - 0.7 < $\mu$ <12.16 + 0.7

11.5< $\mu$ < 13.2

(lower limit $11.5 , upper limit$13.2)

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